Problem: You are given an integer N.

Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.

* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.

Constraints

* 2 \leq N \leq 500

Input

Input is given from Standard Input in the following format:


N


Output

Print your solution in the following format:


a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Output

Print your solution in the following format:


a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Example

Input

2


Output

4 7
23 10

Solution (in CPP):

Solution:
#include <cstdio>
#include <set>
using namespace std;
typedef long long ll;
const int N = 505;
int n, tail;
ll ans[N][N];
int pri[N * N];
bool isNot[N * N];
void Sieve() {
	for (int i = 2; i <= 100000; ++i) {
		if (!isNot[i])
			pri[++tail] = i;
		for (int j = 1; j <= tail && pri[j] * i <= n * n; ++j) {
			isNot[pri[j] * i] = 1;
			if (i % pri[j] == 0)
				continue ;
		}
	}
}
ll gcd(ll x, ll y) { return (!x) ? (y) : gcd(y % x, x); }
ll lcm(ll x, ll y) { return (!x || !y) ? (x | y) : (x / gcd(x, y) * y); }
//这个地方需要先除 gcd，在乘 y，否则 x * y会爆 long long 
int PA[N], PB[N];
int main() {
	scanf("%d", &n);
	if (n == 2) {
		printf("4 7\n23 10\n");
		return 0;
	}
	Sieve();
	for (int i = 1; i <= n; ++i) {
		PA[i] = pri[i];
		PB[i] = pri[i + n];
	}
	int cnt = 0;
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= n; ++j)
			if ((i & 1) == (j & 1)) {
				ll tmp = 1ll * PA[(i + j) / 2] * PB[(i - j) / 2 + (n + 1) / 2];
				ans[i][j] = tmp;
			}
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= n; ++j) {
			if ((i & 1) != (j & 1)) {
				ll tmp = lcm(lcm(ans[i - 1][j], ans[i + 1][j]), lcm(ans[i][j - 1], ans[i][j + 1]));
				ans[i][j] = tmp + 1;
			}
		}
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= n; ++j)
			printf("%lld ", ans[i][j]);
		printf("\n");
	}
	return 0;
}